Hi
chetan2uthis is the OE
The units digit of 2^n is 2, 4, 8, and 6 for n = 1, 2, 3, and 4,
respectively. For n > 4, the units digit of 2^n is equal to that of 2n^−4 .
Thus for every positive integer j the units digit of 2^4j is 6,
and hence 2^2008 has a units digit of 6.
The units digit of 2008^2 is 4.
Therefore the units digit of k is 0,
so the units digit of k^2 is also 0.
Because 2008 is even, both 2008^2 and 2^2008 are
multiples of 4.
Therefore k is a multiple of 4, so the units digit of 2^k is 6, and
the units digit of k^2 + 2^k is also 6
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